Question

if x minus one upon x is equal to 7, evaluate first X square + 1 upon x square

Answer 1

51

Step-by-step explanation:

Given:

$x - \dfrac{1}{x} = 7$

To find:

${x}^{2} + \dfrac{1}{ {x}^{2} }$

Solution:

Just square both side and recover your forgotten Algebra Formula.

$x - \frac{1}{x} = 7 \\ \\ {(x - \frac{1}{x}) }^{2} = {(7)}^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } - 2.x. \frac{1}{x} = 49 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 49 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 49 + 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 51$

Therefore, the required answer is 51.

Important Algebra Formula

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2})$